If $x \veebar y = 3x^{2}+y^{2}$ and $x \bigtriangledown y = xy+2x-y$, find $0 \veebar (-3 \bigtriangledown -3)$.
Answer: First, find $-3 \bigtriangledown -3$ $ -3 \bigtriangledown -3 = (-3)(-3)+(2)(-3)-(-3)$ $ \hphantom{-3 \bigtriangledown -3} = 6$ Now, find $0 \veebar 6$ $ 0 \veebar 6 = 3(0^{2})+6^{2}$ $ \hphantom{0 \veebar 6} = 36$.